Finite square well bound states

In summary, the finite potential well with regions of infinite and finite potential can be solved using the time-independent Schrodinger equation and the continuity conditions of Psi and its derivative. The relation between V0 and the number of bound states can be calculated graphically or numerically, with a possible approximate solution being V0 = (N-1)^2 * (h^2 / 32ma^2).
  • #1
Let's suppose I have a finite potential well: $$
V(x)=
\begin{cases}
\infty,\quad x<0\\
0,\quad 0<x<a\\
V_o,\quad x>a.
\end{cases}
$$

I solved the time-independent Schrodinger equation for each region and after applying the continuity conditions of ##\Psi## and its derivative I ended up with:

$$ \tan(k_1a)=-\frac{k_1}{k_2},$$ where ##k_1=\sqrt{\frac{2mE}{\hbar^2}}## and ##k_2=\sqrt{\frac{2m(V_o-E)}{\hbar^2}}##.

I'm aware of the fact that solutions can only be calculated graphically, but what's the relation between the value of ##V_o## and the bound states? What if I want to find the acceptable values of ##V_o## for the bound states to be ##1,2,3,\dots## or none?
 
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  • #2
andrewtz98 said:
What if I want to find...
andrewtz98 said:
I'm aware of the fact that solutions can only be calculated graphically

Doesn't that answer your question?

There is a relationship between a, m and V0 and the number of bound states. I doubt it has an analytic form, but the way you get it asking that the Nth state be just on the edge of being bound. If I did it right, for large N,

[tex] V_0 \approx (N-1)^2 \frac{h^2}{32ma^2} [/tex]


By the way, graphing is not the only way to find the solutions. You can also do it numerically.
 

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